sodiu cand se introduce in apa formeaza prin reactie NaOH, unde el va fi md
n g b g md g
2Na + 2H2O --> 2NaOH + H2O
2x23 2x18 2x40
stim ca c% = mdx100/ms
unde md = masa de NaOH formata
ms = m.apa initiala - b (masa de appa consumata din reactie) + md
c% = 7,68%
=> 7,68 = mdx100/ms
=> 100md = 7,68(200-b+md)
b = 2x18xmd/2x40 = 0,45md
=> 100md = 7,68(200-0,45md+md)
=> 13,021md = 200+0,55md
=> md = 15,70 g NaOH format
=> n = 2x23xmd/2x40
= 9,026 g Na
